MATH SOLVE

4 months ago

Q:
# I need help with these removable discontinuities.

Accepted Solution

A:

Answer:First problem: a (2,0)Second problem: b. none of these; the answer is (4, 5/3) which is not listed.Third problem: b. none of the above; there are no holes period. Step-by-step explanation:First problem: The hole is going to make both the bottom and the top zero.So I start at the bottom first.[tex]x^2-3x+2=0[/tex]The left hand expression is factorable.Since the coefficient of [tex]x^2[/tex] is 1, you are looking for two numbers that multiply to be 2 and add to be -3.Those numbers are -2 and -1 since (-2)(-1)=2 and -2+(-1)=-3.The factored form of the equation is:[tex](x-2)(x-1)=0[/tex].This means x-2=0 or x-1=0.We have to solve both equations here.x-2=0Add 2 on both sides:x=2x-1=0Add 1 on both sides:x=1Now to determine if x=2 or x=1 is a hole, we have to see if it makes the top 0.If the top is zero when you replace in 2 for x, then x=2 is a hole.If the top is zero when you replace in 1 for x, then x=1 is a hole.Let's do that.[tex]x^2-4x+4[/tex]x=2[tex]2^2-4(2)+4[/tex][tex]4-8+4[/tex][tex]-4+4[/tex][tex]0[/tex]So we have a hole at x=2.[tex]x^2-4x+4[/tex]x=1[tex]1^2-4(1)+4[/tex][tex]1-4+4[/tex][tex]-3+4[/tex][tex]1[/tex]So x=1 is not a hole, it is a vertical asymptote. We know it is a vertical asymptote instead of a hole because the numerator wasn't 0 when we plugged in the x=1.So anyways to find the point for which we have the hole, we will cancel out the factor that makes us have 0/0.So let's factor the denominator now.Since the coefficient of [tex]x^2[/tex] is 1, all we have to do is find two numbers that multiply to be 4 and add up to be -4.Those numbers are -2 and -2 because -2(-2)=4 and -2+(-2)=-4.[tex]f(x)=\frac{(x-2)(x-2)}{(x-2)(x-1)}=\frac{x-2}{x-1}[/tex]So now let's plug in 2 into the simplified version:[tex]f(2)=\frac{2-2}{2-1}=\frac{0}{1}=0[/tex].So the hole is at x=2 and the point for which the hole is at is (2,0).a. (2,0)Problem 2:So these quadratics are the same kind of the ones before. They all have coefficient of [tex]x^2[/tex] being 1.I'm going to start with the factored forms this time:The factored form of [tex]x^2-3x-4[/tex] is [tex](x-4)(x+1)[/tex] because -4(1)=-3 and -4+1=-3.The factored form of [tex]x^2-5x+4[/tex] is [tex](x-4)(x-1)[/tex] because -4(-1)=4 and -4+(-1)=-5.Look at [tex]\frac{(x-4)(x+1)}{(x-4)(x-1)}[/tex].The hole is going to be when you have 0/0.This happens at x=4 because x-4 is 0 when x=4.The hole is at x=4.Let's find the point now. It is (4,something).So let's cancel out the (x-4)'s now.[tex]\frac{x+1}{x-1}[/tex]Plug in x=4 to find the corresponding y:[tex]\frac{4+1}{4-1}{/tex][tex]\frac{5}{3}[/tex]The hole is at (4, 5/3).Third problem:[tex]x^2-4x+4[/tex] has factored form [tex](x-2)(x-2)[/tex] because (-2)(-2)=4 and -2+(-2)=-4.[tex]x^2-5x+4[/tex] has factored form [tex](x-4)(x-1)[/tex] because (-4)(-1)=4 and -4+(-1)=-5.There are no common factors on top and bottom. You aren't going to have a hole. There is no value of x that gives you 0/0.